The value of `9+(16)/(2!)+(27)/(3!)+(42)/(4!)+……infty` is

Options

A. `9e-6`
B. `11e-6`
C. `13e-6`
D. `12e-6`

Explanation

Consider the series
9+16+27+42+….
We observe that the successive differences of the terms of this series are 7,11,15… Clearly these are in A.P so let its `n^(th)` term be
`t_(n)=an^(2)+bn+c`
`rarr t_(1)=a+b+c rarr a+b+c=9`
`t_(2)=4 a+2b+c rarr 4a+2b+c=16`
`t_(3)=9a+3b+c rarr 9a+3b+c=27`
Solving these equation we get
a=2,b=1 and c =6
`therefore t_(n)=2n^(2)+n+6`
Thus we have
`9+(16)/(2!)+(27)/(3!)+(42)/(4!)+...infty`
`=underset(n=1)overset(infty)Sigma(2n^(2)+n+6)/(n!)`
`=2underset(n=1)overset(infty)Sigma(n^(2))/(n!)+underset(n=1)overset(infty)Sigma(n)/(n!)+6 underset(n=1)overset(infty)Sigma(1)/(n!)`
=11e-6